Question

In ∆ABC, ∠B = 35°, ∠C = 65° and the bisector AD of ∠BAC meets BC at D. Then, which of the following is true?
(a) AD > BD > CD
(b) BD > AD > CD
(c) AD > CD > BD
(d) None of these

Answer 1

(b)$BD>AD>CD$

In $∆ABC$, we have:
$\angle B=35°,\angle C=65°$ and the bisector AD of $\angle BAC$ meets BC at D.
Then we have:
$$\begin{gathered} \angle A+\angle B+\angle C=180° \\ \Rightarrow \angle A+35°+65°=180° \\ \Rightarrow \angle A=80° \end{gathered}$$

AD is the angle bisector of $\angle BAC$
∴ $\angle BAD=\angle CAD=40°$

Now, in triangle ABD, we have:
$\angle BAD>\angle ABD$
$\Rightarrow BD>AD$
Also, in triangle ACD, we have:
$\angle ACD>\angle CAD$
$\Rightarrow AD>CD$
$\therefore BD>AD>CD$